A 30.84 mL volume of 0.128 M NaOH is required to reach the phnolphthalein endpoint in the titration of a 5.441 g sample of vinegar. Calculate % acetic acid in the vinegar.
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To calculate the percentage of acetic acid in the vinegar, we can use the concept of stoichiometry and the balanced chemical equation of the reaction between acetic acid and sodium hydroxide.
The balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is:
CH3COOH + NaOH → CH3COONa + H2O
From the equation, we can see that the mole ratio between acetic acid and sodium hydroxide is 1:1. This means that 1 mole of acetic acid reacts with 1 mole of sodium hydroxide.
First, let's calculate the number of moles of sodium hydroxide (NaOH) used:
Moles of NaOH = (Volume of NaOH) × (Molarity of NaOH)
= 30.84 mL × 0.128 mol/L
= 3.95 mmol
Since the mole ratio between acetic acid and sodium hydroxide is 1:1, the number of moles of acetic acid (CH3COOH) present in the vinegar is also 3.95 mmol.
Next, let's calculate the molecular weight of acetic acid (CH3COOH):
Molecular weight of CH3COOH = (12.01 g/mol × 2) + (1.01 g/mol × 4) + (16.00 g/mol) + (1.01 g/mol)
= 60.05 g/mol
Now we can calculate the mass of acetic acid (CH3COOH) present in the vinegar:
Mass of CH3COOH = (Number of moles of CH3COOH) × (Molecular weight of CH3COOH)
= 3.95 mmol × 60.05 g/mol
= 0.237 g
Finally, we can calculate the percentage of acetic acid in the vinegar:
% Acetic Acid = (Mass of CH3COOH / Mass of Vinegar) × 100
= (0.237 g / 5.441 g) × 100
= 4.36%
Therefore, the percentage of acetic acid in the vinegar is approximately 4.36%.